A boy weighing 50 kg finished long jump at a distance of 8 m . Considering that he moved along a parabolic path and his angle of jump is $45^{\circ}$, his initial $K E$ is

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Aicharya · Accepted Answer

$\mathrm{R}=8 \mathrm{~m}, 	heta=45^{\circ}$ $\begin{aligned} & 	herefore \quad R=\frac{u^2 \sin 2 	heta}{g}=\frac{u^2 \sin 90^{\circ}}{g}=\frac{u^2}{9.8} \ & 	herefore \quad u^2=9.8 R=9.8 	imes 8=78.4 \end{aligned}$ $	herefore \quad$ Initial K.E. of the man, $(\mathrm{K} . \mathrm{E})_2=\frac{1}{2} \mathrm{mu}^2=\frac{1}{2} 	imes 50 	imes 78.4=1960 \mathrm{~J}$

A boy weighing 50 kg finished long jump at a distance of 8 m . Considering that he moved along a parabolic…

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